Question
Show that $\left[\begin{array}{rr} {5} & {-1} \\ {6} & {7} \end{array}\right]\left[\begin{array}{ll} {2} & {1} \\ {3} & {4} \end{array}\right] \neq\left[\begin{array}{ll} {2} & {1} \\ {3} & {4} \end{array}\right]\left[\begin{array}{rr} {5} & {-1} \\ {6} & {7} \end{array}\right]$

Answer

L.H.S =$\left[\begin{array}{cc} {5} & {-1} \\ {6} & {7} \end{array}\right]\left[\begin{array}{ll} {2} & {1} \\ {3} & {4} \end{array}\right]$
= $\left[\begin{array}{cc} {5(2)-1(3)} & {5(1)-1(4)} \\ {6(2)+7(3)} & {6(1)+7(4)} \end{array}\right]$
= $\left[\begin{array}{cc} {10-3} & {5-4} \\ {12+21} & {6+28} \end{array}\right]$
= $\left[\begin{array}{cc} {7} & {1} \\ {33} & {34} \end{array}\right]~...(i)$
R.H.S = $\left[\begin{array}{ll} {2} & {1} \\ {3} & {4} \end{array}\right]\left[\begin{array}{ll} {5} & {-1} \\ {6} & {7} \end{array}\right]$
= $\left[\begin{array}{cc} {2(5)+1(6)} & {2(-1)+1(7)} \\ {3(5)+4(6)} & {3(-1)+4(7)} \end{array}\right]$
= $\left[\begin{array}{cc} {10+6} & {-2+7} \\ {15+24} & {-3+28} \end{array}\right]$
= $\left[\begin{array}{ll} {16} & {5} \\ {39} & {25} \end{array}\right]~...(ii)$
Therefore, from (i) and (ii), we get
$\left[\begin{array}{cc} {5} & {-1} \\ {6} & {7} \end{array}\right]\left[\begin{array}{ll} {2} & {1} \\ {3} & {4} \end{array}\right] \neq\left[\begin{array}{ll} {2} & {1} \\ {3} & {4} \end{array}\right]\left[\begin{array}{cc} {5} & {-1} \\ {6} & {7} \end{array}\right]$

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